# How to Multiply Fractions with Variables

When multiplying fractions, I am a fan of factorizing and canceling common factors instead of multiplying and then simplifying.

The reason for this is that it makes the calculations much easier, and I like it when things are easy. By cross-canceling when possible instead of expanding the parentheses, the calculations become simpler and less messy.

Here you will learn to multiply fractions with variables by factorizing and cross-canceling. Beware that you can only cancel factors that are on both sides of the fraction bar.

Rule

### MultiplicationThroughCanceling

1.
Factorize the numerators.
2.
Factorize the denominators.
3.
Cancel common factors.
4.

Example 1

Evaluate $\frac{x+1}{x+2}\cdot \frac{2x+4}{3x+3}$

You factorize and cross-cancel in this way: $\begin{array}{llll}\hfill \frac{x+1}{x+2}\cdot \frac{2x+4}{3x+3}& =\frac{\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+2\right)}\cdot \frac{2\phantom{\rule{-0.17em}{0ex}}\left(x+2\right)}{3\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{(x+1)}}{\text{(x+2)}}\cdot \frac{2\text{(x+2)}}{3\text{(x+1)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Evaluate $\frac{2x+6}{x+3}\cdot \frac{{x}^{2}-9}{x+3}$

You factorize using the third algebraic identity of quadratic expressions, and then you cancel and multiply like this:

$\begin{array}{cc}\frac{2x+6}{x+3}\cdot \frac{{x}^{2}-9}{x+3}& \\ =& \\ \frac{2\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)}& \\ =& \\ \frac{2\text{(x+3)}}{\text{(x+3)}}\cdot \frac{\text{(x+3)}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}{\text{(x+3)}}& \\ =& \\ 2\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)& \\ =& \\ 2x-6& \end{array}$

$\begin{array}{llll}\hfill \frac{2x+6}{x+3}\cdot \frac{{x}^{2}-9}{x+3}& =\frac{2\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2\text{(x+3)}}{\text{(x+3)}}\cdot \frac{\text{(x+3)}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}{\text{(x+3)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Evaluate $\frac{{x}^{2}-1}{x+5}\cdot \frac{{x}^{2}-25}{x-1}$

Use the third algebraic identity of quadratic expressions to factorize, and then cross-cancel and multiply in this manner:

$\begin{array}{cc}\frac{{x}^{2}-1}{x+5}\cdot \frac{{x}^{2}-25}{x-1}& \\ =& \\ \frac{\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)}& \\ =& \\ \frac{\text{(x−1)}\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}{\text{(x+5)}}\cdot \frac{\text{(x+5)}\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}{\text{(x−1)}}& \\ =& \\ \phantom{\rule{-0.17em}{0ex}}\left(x+1\right)\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)& \\ =& \\ {x}^{2}-5x+x-5& \\ =& \\ {x}^{2}-4x-5& \end{array}$

$\begin{array}{llll}\hfill \frac{{x}^{2}-1}{x+5}\cdot \frac{{x}^{2}-25}{x-1}& =\frac{\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x-1\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{(x−1)}\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}{\text{(x+5)}}\cdot \frac{\text{(x+5)}\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}{\text{(x−1)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-5x+x-5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-4x-5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 4

Evaluate $\frac{2{x}^{2}-32}{x-4}\cdot \frac{x+5}{2\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right)}$

You factorize with the help of the third algebraic identity of quadratic expressions, and then you cancel and multiply the fractions in this way:

$\begin{array}{llll}\hfill & \phantom{=}\frac{2{x}^{2}-32}{x-4}\cdot \frac{x+5}{2\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-16\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)}{2\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{2}\phantom{\rule{-0.17em}{0ex}}\left(x+4\right)\text{(x−4)}}{\text{(x−4)}}\cdot \frac{\text{(x+5)}}{\text{2}\text{(x+5)}\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(x+4\right)\cdot \frac{1}{\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{x+4}{x-5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \frac{2{x}^{2}-32}{x-4}\cdot \frac{x+5}{2\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right)}& =\frac{2\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-16\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)}\cdot \frac{\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)}{2\phantom{\rule{-0.17em}{0ex}}\left(x+5\right)\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{2}\phantom{\rule{-0.17em}{0ex}}\left(x+4\right)\text{(x−4)}}{\text{(x−4)}}\cdot \frac{\text{(x+5)}}{\text{2}\text{(x+5)}\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(x+4\right)\cdot \frac{1}{\phantom{\rule{-0.17em}{0ex}}\left(x-5\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{x+4}{x-5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$